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!
Copyright © 2022 by Ian Beardsley!
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Table of Contents!
1.0 Sexagesimal…………………………………………..3!
2.0 The Theory…………………………………………….5!
3.0 Giordano’s Relationship……………………………..13!
4.0 Sexagesimal Mystery Solved……………………….19!
5.0 Discussion…………………………………………….23!
6.0 Proton As Area of One………………………………24!
7.0 Rigorous Formulation of Proton Seconds…………28!
8.0 The Theory…………………………………………….32!
9.0 The Basis………………………………………………43!
10.0 Solar System…………………………………………48!
11.0 Natural Abunndances……………………………… 50!
Appendix 1…………………………………………………53!
Abstract!
There is some indication that 1 second is a natural duration, and I show the duration of 1
second may be tied into Quantum Field Theory because with it the gravitational constant G
outside of quantum mechanics is coupled with Planck’s constant, h intrinsic to quantum
mechanics, and that this predicts the radius of a proton through six-fold symmetry within
experimental errors, as well as the hydrocarbons, the skeletons of biological life chemistry. To
understand the meaning of a second, and why it might be a natural duration I explore its
origins that began with the Sumerians, who mysteriously overnight started civilization in
Mesopotamia. The mathematics they used was sexagesimal and this is why the duration of a
second is what it is. I show both that this duration in our calendar that comes to us from the
Babylonians who used sexagesimal as well, and the Ancient Greeks who did as well, to
reconcile the motion of the moon with that of the Earth, is in the orbital dynamics of the earth
and moon, and in sexagesimal itself. A constant is established as well from Giordano’s
relationship that uses molar mass and the Chandrasekhar limit for a star to not collapse and
become a white dwarf star, that predicts our six-fold symmetry. In the end we show the
diameter of the solar system can be written in astronomical units as the golden ratio over 100
times 360 of a circle and that the radius of a proton is the area of a circle inscribed in an
equilateral triangle divided by pi in terms of the metric system. This results in a theory for the
periodic table of the elements as a mathematical construct. Also we we predict with proton-
seconds the natural relative abundances of the primordial elements hydrogen and helium. We
demonstrate a mathematics for abstract cosmology."
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1.0 Sexagesimal
My book Abstract Cosmology could have easily been called The Book of Six, it was
Buckminster Fuller who said “Nature employs 60 degree coordination”. Sixty degrees are the
degrees of an equilateral triangle. The triangle is the structure that encloses an area with the
least amount of sides. Buckminster Fuller said, systems of triangles are the the only inherently
stable patterns. The regular hexagon, a six-sides polygon tessellates as equilateral, 60 degree,
inherently stable equilateral triangles. It was the scientist Shubnikov who said among the living
organisms the pattern with which we most frequently meet is five-fold symmetry. It is well
known that the physical, like snowflakes, are six-fold symmetry. Thus we should not be
surprised that our Abstract Cosmology is founded on six-fold symmetry. Two and three are the
smallest prime numbers and their product is six. Two times six is twelve, the number most
evenly divisible by whole numbers for its size (it is a so-called abundant number: divisible
evenly by 1,2,3,4,6, their sum is 16 which is greater than twelve itself and, three times six is
eighteen, the cyclical Nature of the periodic table (18 groups). !
At the dawn of civilization, the Sumerians who settled down from wandering, gathering, and
hunting, to invent agriculture and and build ceramic homes, developed the first mathematics,
and it was this sexagesimal (base 60) that passed on to the Babylonians, then ended up with
the Ancient Greeks, who divided not just the hour into 60 minutes of time and the minute into
60 seconds of time, but who divided the sky into hours, minutes, and seconds of arc which
was measured in time counted by the rotation of the Earth, and its orbital period around the
Sun, and the orbital period of the Moon around the Earth.!
We are suggesting nature is founded on six-fold symmetry, six which is constructed by
multiplying together the first two prime numbers, 2 and 3. Two factorial is two, three factorial is
6. Two times six is twelve, the number of months in a year, approximated by our moon’s
approximately 12 orbits around the earth in the time it takes the earth to go around the sun
once. There are four weeks in a lunar month, and!
!
Is the number of seconds in the 24 hour Earth day. The duration of one second comes from
dividing up time and the sky like this in base 60, that came from the Sumerians, Babylonians,
and Ancient Greeks and it is believed they did this because 60 is evenly divisible by!
1,2,3,4,5,6,…12,15,20,30,60,…!
The moon orbits the Earth in approximately 30 rotations of the Earth, hence the 30 day month.
January has 31 days, February 29, March 31, April 30 May 31, June 30, July 31, August 31,
September 30, October 31, November 30, December 31. This averages out to!
!
We divide a circle into 360 units called degrees, which is six squared times ten. As such the
equilateral triangle has each angle equal to 60 degrees. There are approximately 360 earth
rotations in the time it orbits the sun once, Hence in one day the Earth moves through
approximately one degree in one day in its 365 day journey around the Sun. The sidereal
month, the time it takes the moon to return to the same position against the background of the
stars is 27 days 7 hours 43 minutes. We see the power of sexagesimal (base 60) in computing
time with the month divided into days, hours, and seconds where there are 60 minutes in an
1 ⋅ 2 ⋅ 3 ⋅ 4 ⋅ 60
2
= 86,400
31 + 29 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31 + 30 + 31
12
=
367
12
= 30.58333
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hour and 60 seconds in a minute because we can convert this easily into seconds with such a
system as here, the duration of this sidereal month in seconds:!
!
The length of the synodic month, the completion of the phases of the moon, or the time it takes
the moon to return to the same position with respect to the Sun, is 29 days 12 hours 44
minutes 3 seconds. It is a little longer than the sidereal month because the earth has moved
with respect to the sun by the time the moon has come back around to a full orbit. We have:!
!
We looked at the arithmetic mean of the days in the month over a year, Let us look at the
harmonic mean:!
=!
!
Gathering like terms in the denominator:!
=0.225806+0.034482759+0.133333=0.39362!
!
Which is the most frequent value for the days in a month. Now let’s look at the geometric mean
which tempers the various values with one another.!
=30.493!
But perhaps it is more telling to take the mean between the individual days of the month that
we use:!
!
!
0.032258+0.034482759+0.033333=0.10007!
!
!
(24 ⋅ 27 + 7) ⋅ 60
2
+ 43 ⋅ 60
1
+ 0 ⋅ 60
0
= 2,358,000 + 2,580 + 0 = 2360580sec/month
(24 ⋅ 29 + 12) ⋅ 60
2
+ 44 ⋅ 60
1
+ 3 ⋅ 60
0
= 2,548,800 + 2,640 + 3 = 2551443sec/month
12
1
31
+
1
29
+
1
31
+
1
30
+
1
31
+
1
30
+
1
31
+
1
31
+
1
30
+
1
31
+
1
30
+
1
31
12
0.032258 + 0.034482759 + 0.032258 + 0.033333 + 0.032258 + 0.033333 + 0.032258 + 0.032258 + 0.033333 + 0.032258 + 0.033333 + 0.032258
12
0.39362
= 30.486
12
31 ⋅ 29 ⋅ 31 ⋅ 30 ⋅ 31 ⋅ 30 ⋅ 31 ⋅ 31 ⋅ 30 ⋅ 31 ⋅ 30 ⋅ 31
31 + 29 + 30
3
=
90
3
= 30
3
1
31
+
1
29
+
1
30
3
0.10007
= 29.799 ≈ 30
3
31 ⋅ 29 ⋅ 30 = 29.98888 ≈ 30
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We see that the arithmetic mean gives us exactly a 30 day month, hence we always speak
loosely saying a month is 30 days. !
Indeed when Buckminster Fuller speaks of Nature employing 60 degree coordination he is
speaking of the equilateral triangle as the fundamental unit of one for area. That is the six-sided
regular hexagon can be divided into six equilateral triangles, giving it an area of six. Indeed he
takes the equilateral triangle as one because not only does it enclose an area with the fewest
sides, it has the smallest area for its perimeter; a regular hexagon with a perimeter the same
size encloses more area. I would like to say the Ancient Sumerians, Babylonians, and Greeks
made this connection, because in their sexagesimal (base 60) system of notation, one is a
straight line with a triangle on top. See fig 1. Ancient Mesopotamian Cuneiform for writing. It is
not sexagesimal notation in the strict sense, as they have a unique symbol for a tens column,
but it was this sexagesimal structure that lead to the 60 minute hour and 60 second minute, as
well as the angular division projected onto the sky in 60 minutes of arc in a degree and 60
seconds of arc in a minute as a way of mathematically defining time dynamically in terms of the
motions of the Earth and moon.!
2.0 The Theory
Matter is that which has inertia. This means it resists change in position with a force applied to
it. The more of it, the more it resists a force. We understand this from experience, but what is
matter that it has inertia?
In this analogy we are suggesting a proton is a three dimensional bubble embedded in a two
dimensional plane. As such there has to be a normal vector holding the higher dimensional
Fig. 1
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sphere in a lower dimensional space. Thus if we apply a force to to the cross-section of the
sphere in the plane there should be a force countering it proportional to the normal holding it in
a lower dimensional universe. This counter force would be experienced as inertia. (Fig. 2)
Since plank’s constant h is a measure of energy over time where space and time are concerned it
must play a role. Of course the radius of a proton plays a role since squared and multiplied by
it is the surface area of our proton embedded in space. The gravitational constant is force
produced per kilogram over a distance, thus it is a measure of how the surrounding space has an
effect on the proton giving it inertia. The speed of light c has to play a role because it is the
velocity at which events are separated through time. The mass of a proton has to play a role
because it is a measurement of inertia itself. And alas the fine structure constant describes the
degree to which these factors have an effect.
We need offer an interpretation of time, as well. (Fig. 3)
In The Time Machine, a science fiction story by HG Wells, the time traveller describes time as
physical distance, the direction through which the universe is falling at the speed of light, c.
Thus, not only when we move through space do we travel through a distance at a velocity v, but
we travel through a distance t at a velocity c. If we draw the picture and account for that distance
and velocity as well, we arrive at time dilation as given by relativity theory. He writes:
I think that at the time none of us quite believed in the Time Machine. The fact is, the Time
Traveller was one of those men who are too clever to be believed: you never felt that you saw
all around him; you always suspected some subtle reserve, some ingenuity in ambush, behind
his lucid frankness.
4π
Fig. 2
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Since we don’t experience our motion through time as we fall through it with the universe, we
change the sign in . And have the relativistic equation for time dilation:
x = vt
2
0
x
2
+ t
2
= d
2
v
2
t
2
0
c
2
t
2
+
c
2
t
2
0
c
2
t
2
= 1
t
2
0
(
v
2
c
2
+ 1
)
= t
2
t
0
=
t
1 +
v
2
c
2
v
2
/c
2
Fig. 3
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We now show carbon, the core element of life is six-fold symmetric with hydrogen in terms of
the natural constants that characterize space, time, and matter:
(Proton Mass)
(Planck Constant)
(Proton Radius)
(Gravitational Constant)
(light speed)
(Fine Structure Constant)
We find one second gives six protons which is carbon:
Equation 2.1
We find six seconds gives 1 proton is hydrogen:
Equation 2.2
For time t greater than 6 seconds we have fractional protons. For t<6 we the have other
elements.
t
0
=
t
1 −
v
2
c
2
m
P
: 1.67262 × 10
−27
kg
h : 6.62607 × 10
−34
J ⋅ s
r
p
: 0.833 × 10
−15
m
G : 6.67408 × 10
−11
N
m
2
kg
2
c : 299,792,459m /s
α : 1/137
1
α
2
m
p
h 4π r
2
p
Gc
= 6proton ⋅ secon d s = carbon(C )
1
α
2
m
p
h 4π r
2
p
Gc
= 1proton ⋅ 6secon d = hydrogen(H )
!
The fine structure constant squared is the ratio of the
potential energy of an electron in the first circular orbit to
the energy given by the mass of an electron in the Bohr
atom model times the speed of light squared.!
α
2
=
U
e
m
e
c
2
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In that we get one second for carbon and 6 seconds for hydrogen very nearly even, that is
Eq. 2.3
It is suggested that the second is a natural unit. If it is, since it comes from designing a calendar
that reconciles the phases of the moon with the Earth year (12 moons per year, approximately) it
is suggested the unit of a second should be in the Earth-Moon-Sun orbital mechanics. The
translational kinetic energy of the moon and earth are:
1
It turns out:
Eq. 2.4
Where the Lunar Month can be as much as 31 days and is based on the lunar orbital period
(27.32 days). We have
Eq. 2.5
Essentially we have formed a Planck constant, h, for the moon by multiplying its kinetic energy
over the time for the period of its orbit:
If we let the lunar month cancel with moon’s orbital period, we have:
Eq. 2.6
Since
1
6
⋅
1
α
2
m
p
h 4π r
2
p
Gc
= 1.004996352secon d s
K . E . Moon = 3.67E 28J
K . E . Ear th = 2.649E 33J
1secon d ≈
(K Eof Moon)(Lun ar OrbitalPer iod )
LunarMonth
EarthDay
(K Eof Ear th)
31d a ys
(1Ear th Da y)
= 31 ≈ π
3
= 31.006
h = (3.67E 28J )(2.36E6s) = 8.6612E 34J ⋅ s
8.6612E 34J ⋅ s
K Eof Ear th
= 32.696secon ds
1
α
2
m
p
h 4π r
2
p
Gc
≈ 6proton s ⋅
K Eof Moon
K Eof Ear th
⋅ Ear th Da y
See Appendix 1 For Data on Kinetic Energy
1
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is units of mass divided by we can let it cancel with , the mass of a proton, and write:
Eq. 2.7
That is
Explicitly,..
Is proton-seconds. Divide by time and we have a number of protons because it is a mass divided
by the mass of a proton. But these masses can be considered to cancel and leave pure number.
We make a program that looks for close to whole number solutions so we can create a table of
values for problem solving.
By what value would you like to increment?: 0.25
How many values would you like to calculate for t in equation 1 (no more than 100?): 100
24.1199 protons 0.250000 seconds 0.119904 decpart
12.0600 protons 0.500000 seconds 0.059952 decpart
8.0400 protons 0.750000 seconds 0.039968 decpart
6.0300 protons 1.000000 seconds 0.029976 decpart
4.0200 protons 1.500000 seconds 0.019984 decpart
3.0150 protons 2.000000 seconds 0.014988 decpart
2.1927 protons 2.750000 seconds 0.192718 decpart
2.0100 protons 3.000000 seconds 0.009992 decpart
1.2060 protons 5.000000 seconds 0.205995 decpart
1.1486 protons 5.250000 seconds 0.148567 decpart
1.0964 protons 5.500000 seconds 0.096359 decpart
1.0487 protons 5.750000 seconds 0.048691 decpart
1.0050 protons 6.000000 seconds 0.004996 decpart
0.2487 protons 24.250000 seconds 0.248659 decpart
0.2461 protons 24.500000 seconds 0.246121 decpart
0.2436 protons 24.750000 seconds 0.243635 decpart
A very interesting thing here is looking at the values generated by the program, the smallest
integer value 1 second produces 6 protons (carbon) and the largest integer value 6 seconds
produces one proton (hydrogen). Beyond six seconds you have fractional protons, and the rest of
1
1sec
⋅
1
α
2
m
p
h 4π r
2
p
Gc
m
p
m
p
1
α
2
m
p
h 4π r
2
p
Gc
≈ 6 ⋅
K Eof Moon
K Eof Ear th
⋅ Ear th Da y = (6)1.2 secon d s
1
6proton s
⋅
1
α
2
m
p
h 4π r
2
p
Gc
= 1.00secon ds
1
α
2
m
p
h 4π r
2
p
Gc
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the elements heavier than carbon are formed by fractional seconds. These are the hydrocarbons
the backbones of biological chemistry. Here is the code for the program:
#include <stdio.h>
#include <math.h>
int main(int argc, const char * argv[]) {
int n;
float value=0, increment,t=0, p=1.67262E-27, h=6.62607E-34,G=6.67408E-11,
c=299792459,protons[100],r=0.833E-15;
do
{
printf("By what value would you like to increment?: ");
scanf("%f", &increment);
printf("How many values would you like to calculate for t in equation 1 (no more than 100?):
");
scanf("%i", &n);
}
while (n>=101);
{
for (int i=0; i<n;i++)
{
protons[i]=((137*137)/(t*p))*sqrt(h*4*(3.14159)*(r*r)/(G*c));
int intpart=(int)protons[i];
float decpart=protons[i]-intpart;
t=t+increment;
if (decpart<0.25)
{ printf("%.4f protons %f seconds %f decpart \n", protons[i], t-increment, decpart);
}}}}
Carbon equals one second is the radius of a proton:
Equation 2.8
The experimental radius of a proton is:
"
r
p
=
18
3
α
2
m
p
Gc
4πh
= 8.288587 × 10
−16
m = 0.829f m
r
p
= 0.833
±
0.014 f m
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Detour: For equation 2.7 to be perfect Earth day needs to be shorter. A long time ago it was; the Earth loses
energy to the moon. The days become longer by 0.0067 hours per million years. Equation 2.7 is actually 1.2
seconds:!
!
24-20=0.0067t!
t=597 million years!
This was when the earth went through a dramatic change and there was a big explosion of life (The Cambrian).
The dinosaurs went extinct 65 million years ago giving small mammals a chance to evolve paving the way for
humans.!
24-x=0.0067t!
x=23.5645 hours!
We say 20 hours + 3 hours is 0 hours + 3 hours since 20 hours is the zero of our cosmic calendar:!
!
What is the next term?!
24hrs!
20+3+0.57735+0.4714=24hours!
Which brings us to today. Indeed even though the earth day gets longer because of the moon the earth year
does not for all practical purposes, however we can consider the solar system today as the end result of the
evolution of an accretion disc into the current distribution of planets, which is as it is because the planets don’t
effect one another (pull each other out of their orbits). But if we must define a state for the end result of the
solar system in its perfected state perhaps we can consider its state at a zero-point of the Cambrian.
24h ou rs
1.2
= 20h ours
3cos(0
∘
) +
2
3
cos(30
∘
) = di n os a ur − e x t i n ct i on =
3h our s +
3
3
h our s
20hr s + 3h r s +
3
3
hr s +
2
3
=
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3.0 Giordano’s Relationship We now formulate what I call Giordano’s Relationship: Warren
Giordano writes in his paper The Fine Structure Constant And The Gravitational Constant: Keys
To The Substance Of The Fabric Of Space, March 21, 2019:
In 1980, the author had compiled a series of notes analyzing Einstein’s geometric to kinematic
equations, along with an observation that multiplying Planck’s constant ‘ ’ by ‘ ’, where ‘ ’
is the Fine Structure Constant, and multiplying by yielded Newton’s gravitational
constant numerically, but neglecting any units.
Let’s do that
(6.62607E-34Js)(1+1/137)(1E23)=6.6744E-11 Js
And it works, G is:
G=6.67408E-11 N(m2/kg2)
Let us reformulate this as:
Equation 3.1
Where and H=1 gram/atom
Because for hydrogen 1 proton is molar mass 1 gram, for carbon 6 protons is 6 grams and so on
for 6E23 atoms per gram. Thus,…
Since grams and atom cancel we can work in grams even though our equations are in kilograms.
Let us not write H, since formally it is grams per mole of hydrogen but write
We have:
Or,…
Equation 3.2
Where
Equation 3.3
Let us say we were to consider Any Element say carbon . Then in general
h
1 + α
α
10
23
h
(1 + α)
G
N
A
H = 6.0003
kg
2
⋅ s
m
N
A
= Avaga dro′ s − Nu m ber = 6.02E 23atom s /gram
N
A
H = 6.02 E 23
atom s
gram
⋅
1gram
atom
= 6.02E 23
ℍ = 1
gram
atom
h
(1 + α)
G
⋅ N
A
ℍ = 6.0003kg
2
⋅
s
m
h(1 + α)N
A
ℍ = 6Gx
x = 1.00kg
2
s
m
𝔼
ℂ
of 14 54
Equation 3.4
We have
and
Because there are six grams of protons in carbon which has 6 protons and 6 neutrons and a
molar mass of 12. We have
12-6=6 grams of protons in the 12 grams of protons and neutrons. Thus
And it follows that
We see in general since the atomic number Z is the number of protons in an atom that in general
this holds for all elements because
And,
Therefore we always have:
Equation 3.5
This works nicely because we formulated molar mass nicely; we said element one (hydrogen)
which is one proton and one electron has one gram for a mole of atoms. Historically this was
done because we chose carbon (element six) to have 12 grams per mole, and determined what
the mole was such that it would hold. The reason this works is that hydrogen is one proton and
has no neutrons, but carbon has twelve neutrons but since hydrogen doesn’t have any neutrons,
and the neutron has the same mass as the proton, and our theory makes use only of protons (in
this instance of its formulation) equation 3.3
Comes out to have x equal to 1.00 (nearly) even. It is at this moment that we point out, because
it is important, that in equation 3.5
h
(1 + α)
G
⋅ N
A
𝔼 = 6.0003kg
2
⋅
s
m
ℂ =
6grams
6proton s
N
A
=
6(6E 23pr oton s)
6grams
N
A
ℂ = 6E 23
h
(1 + α)
G
⋅ N
A
ℂ = 6.0003kg
2
⋅
s
m
𝔼
N
A
=
Z ⋅ 6E 23proton s
Z ⋅ gr a m s
𝔼 =
Z ⋅ gr a m s
Z ⋅ proton s
N
A
⋅ 𝔼 = 6E 23
x = 1.00kg
2
s
m
N
A
⋅ 𝔼 = 6E 23
of 15 54
is not molar mass, and that is a variable determined by ; it is the number of a mole of
atoms multiplied by the number of protons in . The reason we point this out, though it may
already be clear, is we wish to find the physical theory behind it. That is we need to find the
physical explanation for equation 3.4
It is the integer 6 to 3 ten thousandths. Which classifies it as interesting because since it is in
kilograms, seconds, and meters, it may mean these units of measurement have some kind of a
meaning. We can in fact write it:
We know that
The fine structure constant squared is the ratio of the potential energy of an electron in the first
circular orbit to the energy given by the mass of an electron in the Bohr model times the speed of
light squared. To begin our search for the meaning of equation 1.4 we convert x, the factor of
1.00 to astronomical units, years, and solar masses, as these are connected to the orbit
of earth as it relates to the sun. We have:
=
We can now write
Eq 3.6.
This unit of AU/year is very interesting. It is not , which would be the Earth’s orbital
velocity, but is a velocity given by the earth orbital radius to its orbital period, which is quantum
mechanical in nature. It relates to earth as as a state, as we have with atoms, a number. We
multiply both sides by and we have earth velocity on the left and the units stay the same on
the right. But what we will do is return to the form in kg-m-s and leave it as an equation but put
in the Earth mean orbital velocity which is 29.79km/s (Zombeck, Martin V. 1982). We get:
𝔼
N
A
𝔼
𝔼
h
(1 + α)
G
⋅ N
A
𝔼 = 6.0003kg
2
⋅
s
m
h
(1 + α)
G
⋅ N
A
𝔼 = 6.000kg
2
⋅
s
m
α
2
=
U
e
m
e
c
2
kg
2
⋅
s
m
kg
2
1
⋅
s
m
⋅
(1.98847E 30)
2
M
2
⊙
kg
2
1.4959787E11m
AU
⋅
year
3.154E 7s
1.8754341E64
M
2
⊙
⋅ year
AU
h
(1 + α)
G
⋅ N
A
𝔼
AU
year
= 8.2172E 32M
⊙
2π AU/year
4π
2
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Eq. 3.7
This brings up an interesting question: while we have masses characteristic of the microcosmos
like protons, and masses characteristic of the macrocosmos, like the minimum mass for a star to
become a neutron star as opposed to a white dwarf after she novas (The Chandrasekhar limit)
which is 1.44 solar masses, we do not have a characteristic mass of the intermediary world where
we exist, a truck weighs several tons and tennis ball maybe around a hundred grams. To find
that mass let us take the geometric mean between the mass of a proton and the mass of 1.44
solar masses. We could take the average, or the harmonic mean, but the geometric mean is the
squaring of the proportions, it is the side of a square with the area equal to the area of the
rectangle with these proportions as its sides. We have:
We multiply this by 1.44 to get 2.8634E30kg. The mass of a proton is .
We have the intermediary mass is:
Eq. 3.8
All we really need to do now is divide 2.7 by 2.8 and we get an even number that is the six of our
six-fold symmetry.
Eq. 3.9
The six of our six-fold symmetry.
We have something very interesting here. We have
This is:
Equation 3.10
Where k is a constant, given
Equation 3.11
We can take the velocity of earth as being 30,000 m/s by rounding it. We have
h
(1 + α)
G
⋅ N
A
𝔼 ⋅ v
e
= 422.787kg
M
⊙
= 1.98847E 30kg
m
p
= 1.67262E − 27kg
m
i
= (2.8634E 30)(1.67262E − 27) = 69.205kg
1
m
i
h
(1 + α)
G
⋅ N
A
𝔼 ⋅ v
e
= 6.1092 ≈ 6
1
69.205kg
6kg
2
s
m
v
e
= 6
k v
e
= 6
k =
1
800
s
m
30,000
800
= 37
1
2
37.5 = 6.123734357
of 17 54
Using , we write
1
6
⋅
1
α
2
m
p
h 4π r
2
p
Gc
= 1.004996352secon d s
k v
e
= 6
1
α
2
m
p
h 4π r
2
p
Gck v
e
= 1proton ⋅ secon d
(K . E . Moon)(Ear th Da y)
(K . E . Ear th)
≈ 1secon d
The second comes from the Ancient Greeks
dividing a minute into 60 seconds and an hour
into 60 minutes because they used sexagesimal
counting (base 60). Probably because 60 is
evenly divisible by:!
1,2,3,4,5,6,…12,15,20.30,60,…!
Mercury= !
Mars= !
Saturn=
60
30
12
Fig. 3
of 18 54
Eq. 3.12
The primordial element from which all others were made.
We can explicitly write the constant k:
Equation 3.13
It was the Indian physicist Chandrasekhar who found the limit in mass for which a white dwarf
will not have its gravity overcome the degeneracy pressure and collapse. The non-relativistic
equation is:
Equation 3.14
Let us approximate 0.77 with 3/4. Since we have our constant
And
Then
Equation 3.15
Since our constant k in terms the Chandrasekhar limit is
Equation 3.16
1
α
2
m
p
h 4π r
2
p
Gck v
e
(K . E . Ear th)
(K . E . Moon)(Ear th Da y)
= 1proton
1
α
2
m
p
h 4π r
2
p
Gc
(K . E . Ear th)
(K . E . Moon)(Ear th Da y)
= 6proton s
1
α
2
m
p
h 4π r
2
p
Gck v
e
(K . E . Ear th)
(K . E . Moon)(Ear th Da y)
= ℍ
ℍ = hydr ogen
k =
1
m
2
i
h
(1 + α)
G
⋅ N
A
𝔼
M ≤ 0.77
c
3
ℏ
3
G
3
N
m
4
p
= 1.41 ⊙
k =
1
m
2
i
h
1 + α
G
⋅ N
A
𝔼
m
i
= Mm
p
m
i
=
3
2
c
3
ℏ
3
G
3
m
2
p
1/2
ℏ = h /2π
k =
4
3
m
p
8π
3
G
c
3
h
(1 + α) ⋅ N
A
𝔼
of 19 54
4.0 Sexagesimal Mystery Solved We now search for the mystery hidden in the sexagesimal
solar system. We begin with the Earth rotates through 15 degrees in one hour:!
!
The distance then that the earth equatorial surface rotates through is where r is the
earth radius and theta is in radians.!
!
Now we look at how many degrees through which the earth rotates in 1 minute:!
!
!
!
For seconds…!
!
!
!
Now let’s look at the same for the distance the earth moves around the Sun in an hour, a
minute, and a second as well…!
!
!
!
!
!
!
And finally…!
360
24
= 15
s = r θ
s = (6,378.14)(0.2618rad ) = 1669.8k m /hr
24(60) = 1440min
360
1440
= 0.25
∘
= 0.0043633ra d
s = (6,378.14)(0.00436) = 27.83k m /min
24(60)(60) = 86,400s
360
∘
86,400
= 0.004167
∘
= 0.00007272ra d
s = (6,378,14)(0.00007272) = 0.464k m /s
(365.25)(24) = 8766hr /yr
360
8766
= 0.041
∘
= 0.00071678ra d
(149,598,000)(0.000716768) = 107,227k m /hr
(365.25)(24)(60) = 525960
360
525960
= = 0.000684463
∘
= 0.000011946ra d
s = (149,598,000)(0.000011946) = 1,787.1k m /min
of 20 54
!
!
!
We have the following table:!
As we can see I am in good agreement with Martin Zombek, Handbook of Space Astronomy
and Astrophysics which provides data. Notice 27.83 km/min in earth rotation is approximately
the 29.786 km/sec in earth orbit. That is a valuable clue. Now let us consider!
Earth: 1 AU=149,598,000km (average earth-sun separation).!
!
The earth rotates through about one degree a day. !
!
Earth Orbit:!
!
!
Earth Rotation: r=6,378.14km!
!
(365.25)(24)(60)(60) = 31557600
360
31557600
= 0.000011408
∘
= 0.000000199ra d
(149,598,000)(0.000000199) = 29.786k m /sec
360
365.25
= 0.9856
deg
day
≈ 1
deg
day
360
∘
= 2π radia n s = 6.283
ra d
year
0.9856
∘
= 0.017202424
ra d
day
(149,598,000)(0.017202424) = 2,573,448.201
k m
day
360
∘
1day
= 6.283
ra d
day
of 21 54
!
Moon: sidereal month =27.83 days, r=405,400 km!
!
!
!
!
The points to be made in this exploration!
1. So the earth goes through 1 degree a day and the moon 1 kilometer per second.!
2. Earth rotates through 27.83 km/min at its equatorial surface and orbits through 29.286 km/s
around the sun. These two are close to the same.!
3. The synodic month is 29.53 days approximately equals 29.786 km/s. The sidereal month is
27.32 days is approximately 27.83 km/min. !
With that information we have this mystery of sexagesimal in the earth-moon-sun orbital
parameters, solved. We see we can make the following equation in table form:!
!
s = r θ = (6,378.14)(6.283) = 40,075.0
k m
day
(27.83)(24) = 667.92
hours
month
360
667.92
= 0.529
∘
= 0.0094
ra d
hr
(405,400k m)(0.0094ra d /hr) = 3,810.76k m /hr
(3,810.75k m /hr)(hr /60min) = 63.5k m /min = 1.0585k m /s
of 22 54
!
1 astronomical unit (AU) =149,598,000km is the average earth distance from the sun in its
nearly circular orbit.!
!
Is the diameter of the earth orbit 2AU. We define our variables:!
Earth orbits: !
Earth rotates: !
Earth orbits: !
Moon orbits: !
Earth completes a 360 degree orbit yields:!
!
Where on the right it is in radians and is the radius of the Earth’s orbit. We have!
!
This is (0.00618)360=2.225!
0.00618 is !
Where is the inverse of the golden ratio. .!
We have:!
!
86,400
k m
deg
⋅ 360
∘
= 311,040,000k m
311,040,000k m
149,598,000k m /AU
= 2.079AU
v
e
= 29.786
k m
s
ω
e
=
27.83k m
min
=
27.83k m
min
⋅
min
60sec
= 0.4638
k m
sec
θ
e
=
1day
deg
=
(24)(60)(60)
deg
= 14,400
sec
deg
v
m
=
1k m
sec
v
e
⋅ θ
e
⋅ v
m
ω
e
⋅
360
∘
1AU
= 2π
v
e
⋅ θ
e
⋅ v
m
ω
e
⋅ r
e
r
e
= 1AU
(29.786)(14,400)(1k m /s)
(0.4638)(149,598,000)
360 = 2.225AU
ϕ
100
ϕ
ϕ = 0.618 =
1
Φ
ϕ
100
360
∘
= 2AU
of 23 54
!
We see in fig. 1 of section 1.0 that the Babylonian symbols in sexagesimal also have a cycle of
base 10, that which we use today. It is defined:!
!
5.0 Discussion
The duration of a second is not constant, we have to slightly modify and adjust it because it is
measured by the rate the earth spins, which slows down very slowly, the Earth loses energy to
the moon. The days become longer by 0.0067 hours per million years. I propose we can define
the second in terms of the natural constants with my equation for proton seconds:!
!
Indeed the second can be connected to the Natural constants. Even though the solar system
has been changing by small amounts over time, its current state is the culmination of its
evolution which gives a distribution of the planets such that they interfere with one another
minimally. If they did interfere with one another significantly, they wouldn’t be there, they would
have pulled each other from their current periods and altered climate on earth so significantly
that life could not have had the chance to evolve over the billions of years or even from their
orbits completely. The second can be connected to the natural constants because the rotation
of the planets, and their orbital periods was determined by the rotational properties of the
accretion disc from which they formed, and that in turn has to be determined in large part by
the masses and radii of the protons from which they formed from this accretion disc, and, the
mass and radius of the proton is determined by the natural constants according to my equation
for proton seconds:!
!
Which is evaluated at t=1 second. And as well we see the kinetic energies of the Earth and
Moon in their orbit which resulted from the rotation of the protoplanetary accretion disc gives
about one second:!
!
It is actually 1.2 seconds which corresponds to the Earth day as it was 597 million years ago,
which was the Cambrian when there was a giant explosion of life into the vast diversification of
species.!
I am not saying we intentionally devised the duration of a second to be connected to the
natural constants, but rather we did because we did attempt to make sense of the rotation and
orbit of the Earth in terms of the phases of the moon and that was done with a choice for a
counting system (sexagesimal, base 60) which is so divisible evenly by important numbers that
coincide with the periods of the moon and the Earth, as I showed in my equations that give the
diameter of the Earth orbit !
10
0
= 1,10
1
= 1,10
2
= 100,...
1
6
⋅
1
α
2
m
p
h4π r
2
p
Gc
= 1.004996352seconds
r
p
=
18
3
α
2
m
p
Gc
4πh
= 8.288587 × 10
−16
m = 0.829f m
(K . E . Moon)(Ear th Da y)
(K . E . Ear th)
≈ 1secon d
of 24 54
!
!
360 is approximately the number of days in a year where a day is one turn of the Earth and we
wisely divide the circle into 360 units (degrees) which is 60 squared of our 60 degree
coordination based upon our six-fold symmetry at the basis of the physical.!
6.0 Proton As Area of One
We showed that a second is described by the natural constants:!
!
And that the diameter of the Earth orbit can nicely be written as the golden ratio reduced by
the 100 of our base ten system of counting times the 360 degrees that make a circle:!
!
Because!
!
Now we seek to show that the radius of a proton is the area of a circle inscribed in an
equilateral triangle in terms of the metric system. We have equation 3.10!
For Earth!
For Jupiter!
For Uranus!
If the velocity of Earth is taken as 30,000 km/s it works better than its actual 29,790 km/s for
these planets to be included in our equations making k=0.0012 and 1/k=833.333…=30,000/36.
We also want to do this because the experimental value of the radius of a proton in one of the
more recent measurements is:!
And we have predicted:
v
e
⋅ θ
e
⋅ v
m
ω
e
360
∘
= D
ϕ
100
360 = 2AU
1
6
⋅
1
α
2
m
p
h4π r
2
p
Gc
= 1.004996352seconds
ϕ
100
360
∘
= 2AU
v
e
⋅ θ
e
⋅ v
m
ω
e
360
∘
= D
k v
e
= 6
k v
J
= 4
k v
u
= 3
r
p
= 0.833
±
0.014 f m
of 25 54
Which is evaluated at one second where one second is given by
We now consider the hydrocarbons where the smallest integer value 1 second in our
equation for proton seconds produces 6 protons (carbon) and the largest integer value 6
seconds produces one proton (hydrogen). (Beyond six seconds you have fractional protons,
and the rest of the elements heavier than carbon are formed by fractional seconds.) Thus we
take the six-sided regular hexagon with six closest packed circles as protons to have a side of
one second to be carbon and the six-sided regular hexagon with a circle inscribed in it and
radii adding up to six seconds to be hydrogen:!
The apothem, a, for the equilateral triangle is
And the area of the circle inscribed in it is then!
!
r
p
=
18
3
α
2
m
p
Gc
4πh
= 8.288587 × 10
−16
m = 0.829f m
(K . E . Moon)(Ear th Day)
(K . E . Ear th)
≈ 1second
a =
3
6
A = π r
2
of 26 54
Where r=a. This gives!
and !
But the radius of a proton is . Therefore
In femtometers (fm). See the brainstorm on the next page…
A =
π
12
A
π
=
1
12
= 0.08333...
r
p
= 0.833
±
0.014 f m
A
10π
= r
p
of 27 54
!
of 28 54
7.0 Rigorous Formulation of Proton Seconds
We can actually formulate this differently than we have. We had
But if t1 is not necessarily 1 second, and t6 is not necessarily six seconds, but rather t1 and t2 are
lower and upper limits in an integral, then we have:
Equation 7.1
This Equation is the generalized equation we can use for solving problems.
Essentially we can rigorously formulate the notion of proton-seconds by considering
Equation 7.2
Is protons-seconds squared where current density is and ( can also be
). We say
Equation 7.3
Keeping in mind q is not charge (coulombs) but a number of charges times seconds, here a
number of protons. It is
Equation 7.4
Dividing Equation 7.2 through by by t:
Equation 7.5
Which is proton-seconds. Dividing through by t again:
1
t
1
1
α
2
m
p
h 4π r
2
p
Gc
= 6proton s
1
t
6
1
α
2
m
p
h 4π r
2
p
Gc
= 1proton
1
α
2
m
p
h 4π r
2
p
Gc
∫
t
2
t
1
1
t
2
dt = ℕ
∫
t
qdt = t
2
∬
S
ρ(x, y, z)d x d y
J = ρ
v
ρ = Q /m
3
ρ
Q /m
2
Q =
∫
V
ρd V
ℕ =
1
α
2
m
p
h 4π r
2
p
Gc
1
α
2
m
p
h 4π r
2
p
Gc
∫
t
dt
t
= t
∬
S
ρ(x, y, z)d x d y
of 29 54
Equation 7.6
We see that if where and then J is I/m2 (current per square meter)
is analogous to amperes per per square meter which are coulombs per second through a surface.
Thus we are looking at a number of protons per second through a surface. Thus we write:
Is carbon where 0.5 seconds is magnesium (Mg) from the values of time corresponding to
protons in the output from our program and 1.0 seconds is carbon (C). We see we have the
following theorem:
Equation 7.7
So as an example,…
Is fluorine (F). Divide by xy with x=y=1 and we have current density. And multiply by 1 second
which is carbon and we have protons per square meter.
1
α
2
m
p
h 4π r
2
p
Gc
∫
t
dt
t
2
= proton s
J = ρ
v
ρ = Q /m
3
v = m /s
1
α
2
m
p
h 4π r
2
p
Gc
∫
t
C
t
Mg
dt
t
2
= 6
∫
1.0
0.5
t
−2
dt = − 6(1 − 2) = 6
1
α
2
m
p
h 4π r
2
p
Gc
∫
t
dt
t
3
=
∬
S
J ⋅ d
S
1
α
2
m
p
h 4π r
2
p
Gc
∫
1.0
0.5
dt
t
3
=
∬
S
J ⋅ d
S = − 3
(
1 −
1
0.25
)
= 9
proton s
secon d
J(x, y, z) = (0,0, J ) = − J
k
d
S = d x d y
k
J ⋅ d
S = (0,0, J ) ⋅ (0,0, d x d y) = − Jd x d y
of 30 54
We are now equip to do computations in proton-seconds. We use equation 7.6 from two to
three, the smallest prime numbers that multiply to make six-fold symmetry in our hexagonal
proton that we found described its radius (My feeling is we introduce the factor of 2 because
carbon is 6 protons +6 neutrons and 2 times 6 is twelve):
But we already know this is the radius of a proton in fm, it is as if fm is a natural length, as well
as the second we have been using. 2 seconds in our program is 3 protons (Lithium Li) and 3
seconds is two protons is helium, He:
By what value would you like to increment?: 0.01
How many values would you like to calculate for t in equation 1 (no more than 100?): 100
86.1425 protons 0.070000 seconds 0.142517 decpart
50.2498 protons 0.120000 seconds 0.249805 decpart
43.0713 protons 0.140000 seconds 0.071259 decpart
40.1998 protons 0.150000 seconds 0.199841 decpart
30.1499 protons 0.200000 seconds 0.149876 decpart
26.2173 protons 0.230000 seconds 0.217283 decpart
25.1249 protons 0.240000 seconds 0.124893 decpart
24.1199 protons 0.250000 seconds 0.119900 decpart
23.1922 protons 0.260000 seconds 0.192213 decpart
20.0999 protons 0.300000 seconds 0.099920 decpart
17.2285 protons 0.350000 seconds 0.228504 decpart
15.0749 protons 0.400000 seconds 0.074944 decpart
14.0232 protons 0.430000 seconds 0.023204 decpart
13.1086 protons 0.460000 seconds 0.108647 decpart
12.0600 protons 0.500000 seconds 0.059957 decpart
11.1666 protons 0.540000 seconds 0.166627 decpart
2
α
2
m
p
h 4π r
2
p
Gc
∫
t
dt
t
2
= proton s
2
α
2
m
p
h 4π r
2
p
Gc
∫
3
2
dt
t
3
= 0.833proton s /secon d
of 31 54
10.2203 protons 0.590000 seconds 0.220302 decpart
10.0500 protons 0.600000 seconds 0.049964 decpart
9.1363 protons 0.660000 seconds 0.136332 decpart
8.1486 protons 0.740000 seconds 0.148621 decpart
8.0400 protons 0.750000 seconds 0.039971 decpart
7.1785 protons 0.839999 seconds 0.178546 decpart
7.0941 protons 0.849999 seconds 0.094093 decpart
7.0116 protons 0.859999 seconds 0.011603 decpart
6.2165 protons 0.969999 seconds 0.216474 decpart
6.1530 protons 0.979999 seconds 0.153040 decpart
6.0909 protons 0.989999 seconds 0.090889 decpart
By what value would you like to increment?: 0.25
How many values would you like to calculate for t in equation 1 (no more than 100?): 100
24.1199 protons 0.250000 seconds 0.119904 decpart
12.0600 protons 0.500000 seconds 0.059952 decpart
8.0400 protons 0.750000 seconds 0.039968 decpart
6.0300 protons 1.000000 seconds 0.029976 decpart
4.0200 protons 1.500000 seconds 0.019984 decpart
3.0150 protons 2.000000 seconds 0.014988 decpart
2.1927 protons 2.750000 seconds 0.192718 decpart
2.0100 protons 3.000000 seconds 0.009992 decpart
1.2060 protons 5.000000 seconds 0.205995 decpart
1.1486 protons 5.250000 seconds 0.148567 decpart
1.0964 protons 5.500000 seconds 0.096359 decpart
1.0487 protons 5.750000 seconds 0.048691 decpart
1.0050 protons 6.000000 seconds 0.004996 decpart
0.2487 protons 24.250000 seconds 0.248659 decpart
0.2461 protons 24.500000 seconds 0.246121 decpart
0.2436 protons 24.750000 seconds 0.243635 decpart
Thus we write
Equation 7.8.
Thus we suggest the radius of a proton of a proton (0.833protons) is the proton radius .
Now we integrate from Aluminum is 13 protons=0.372 seconds to phosphorus is 15 protons =
0.396 seconds which is to integrate across silicon:
By what value would you like to increment?: 0.006
How many values would you like to calculate for t in equation 1 (no more than 100?): 100
251.2490 protons 0.024000 seconds 0.248978 decpart
59.1174 protons 0.102000 seconds 0.117416 decpart
50.2498 protons 0.120000 seconds 0.249805 decpart
40.1998 protons 0.150000 seconds 0.199844 decpart
37.2221 protons 0.162000 seconds 0.222080 decpart
27.1621 protons 0.222000 seconds 0.162058 decpart
25.1249 protons 0.240000 seconds 0.124907 decpart
20.0999 protons 0.300000 seconds 0.099916 decpart
17.0338 protons 0.354000 seconds 0.033823 decpart
16.2096 protons 0.372000 seconds 0.209604 decpart
15.2272 protons 0.396000 seconds 0.227202 decpart
2
α
2
m
p
h 4π r
2
p
Gc
∫
t
He
t
Li
dt
t
3
= 0.833proton s /secon d = r
proton
r
proton
of 32 54
14.1549 protons 0.426000 seconds 0.154862 decpart
13.2236 protons 0.456000 seconds 0.223620 decpart
13.0519 protons 0.462000 seconds 0.051885 decpart
12.1084 protons 0.498000 seconds 0.108375 decpart
11.1666 protons 0.540000 seconds 0.166615 decpart
11.0439 protons 0.546000 seconds 0.043906 decpart
10.1515 protons 0.594000 seconds 0.151471 decpart
Equation 7.9.
Silicon can be doped with phosphorus to make negative (n-type) silicon that semi-conducts thus
enabling the construction of logic circuits that you can use to make computing machines. But
this must be joined with positive (p-type) silicon which usually uses boron, but boron is in the
same group as aluminum, just above it. This results in a theory for AI elements as mathematical
constructs, that we will go into now.
8.0 The Theory
Above we see the artificial intelligence (AI) elements pulled out of the periodic table of the
elements. As you see we can make a 3 by 3 matrix of them and an AI periodic table. Silicon and
germanium are in group 14 meaning they have 4 valence electrons and want 4 for more to attain
noble gas electron configuration. If we dope Si with B from group 13 it gets three of the four
electrons and thus has a deficiency becoming positive type silicon and thus conducts. If we dope
the Si with P from group 15 it has an extra electron and thus conducts as well. If we join the two
types of silicon we have a semiconductor for making diodes and transistors from which we can
make logic circuits for AI.
As you can see doping agents As and Ga are on either side of Ge, and doping agent P is to the
right of Si but doping agent B is not directly to the left, aluminum Al is. This becomes important.
I call (As-Ga) the differential across Ge, and (P-Al) the differential across Si and call Al a dummy
in the differential because boron B is actually used to make positive type silicon.
That the AI elements make a three by three matrix they can be organized with the letter E with
subscripts that tell what element it is and it properties, I have done this:
2
α
2
m
p
h4πr
2
p
Gc
∫
t
Al
t
P
dt
t
3
= − 6(6.377 − 7.22) = 5protons /secon d = boron
of 33 54
Thus E24 is in the second row and has 4 valence electrons making it silicon (Si), E14 is in the
first row and has 4 valence electrons making it carbon (C). I believe that the AI elements can be
organized in a 3 by 3 matrix makes them pivotal to structure in the Universe because we live in
three dimensional space so the mechanics of the realm we experience are described by such a
matrix, for example the cross product. Hence this paper where I show AI and biological life are
mathematical constructs and described in terms of one another.
We see, if we include the two biological elements in the matrix (E14) and and (E15) which are
carbon and nitrogen respectively, there is every reason to proceed with this paper if the idea is to
show not only are the AI elements and biological elements mathematical constructs, they are
described in terms of one another. We see this because the first row is ( B, C, N) and these
happen to be the only elements that are not core AI elements in the matrix, except boron (B)
which is out of place, and aluminum (Al) as we will see if a dummy representative makes for a
mathematical construct, the harmonic mean. Which means we have proved our case because the
first row if we take the cross product between the second and third rows are, its respective unit
vectors for the components meaning they describe them!
The Computation
E
13
E
14
E
15
E
23
E
24
E
25
E
33
E
34
E
35
A = (Al, Si, P )
B = (Ga, G e, As)
A ×
B =
B
C
N
Al Si P
Ga Ge As
= (Si ⋅ As − P ⋅ G e)
B + (P ⋅ Ga − Al ⋅ As)
C + (Al ⋅ G e − Si ⋅ G a)
N
A ×
B = − 145
B + 138
C + 1.3924
N
A = 26.98
2
+ 28.09
2
+ 30.97
2
= 50g/mol
B = 69.72
2
+ 72.64
2
+ 74.92
2
= 126g/mol
A ⋅
B = A Bcosθ
cosθ =
6241
6300
= 0.99
θ = 8
∘
A ×
B = A Bsinθ = (50)(126)sin8
∘
= 877.79
of 34 54
And silicon (Si) is at the center of our AI periodic table of the elements. We see the biological
elements C and N being the unit vectors are multiplied by the AI elements, meaning they
describe them. But we have to ask; Why does the first row have boron in it which is not a core
biological element, but is a core AI element? The answer is that boron is the one AI element that
is out of place, that is, aluminum is in its place. But we see this has a dynamic function.
The Dynamic Function
The primary elements of artificial intelligence (AI) used to make diodes and transistors, silicon
(Si) and germanium (Ge) doped with boron (B) and phosphorus (P) or gallium (Ga) and arsenic
(As) have an asymmetry due to boron. Silicon and germanium are in group 14 like carbon (C)
and as such have 4 valence electrons. Thus to have positive type silicon and germanium, they
need doping agents from group 13 (three valence electrons) like boron and gallium, and to have
negative type silicon and germanium they need doping agents from group 15 like phosphorus
and arsenic. But where gallium and arsenic are in the same period as germanium, boron is in a
different period than silicon (period 2) while phosphorus is not (period 3). Thus aluminum (Al)
is in boron’s place. This results in an interesting equation.
Equation 8.1.
The differential across germanium crossed with silicon plus the differential across silicon
crossed with germanium normalized by the product between silicon and germanium is equal to
the boron divided by the average between the germanium and the silicon. The equation has
nearly 100% accuracy (note: using an older value for Ge here, it is now 72.64 but that makes the
equation have a higher accuracy):
Due to an asymmetry in the periodic table of the elements due to boron we have the
harmonic mean between the semiconductor elements (by molar mass):
Equation 8.2
This is Stokes Theorem if we approximate the harmonic mean with the arithmetic mean:
877.79 = 29.6g /m ol ≈ Si = 28.09g/mol
Si(A s − Ga) + Ge(P − Al )
SiGe
=
2B
Ge + Si
28.09(74.92 − 69.72) + 72.61(30.97 − 26.98)
(28.09)(72.61)
=
2(10.81)
(72.61 + 28.09)
0.213658912 = 0.21469712
0.213658912
0.21469712
= 0.995
Si
B
(A s − Ga) +
Ge
B
(P − Al ) =
2SiGe
Si + G e
∫
S
( ∇ × u ) ⋅ d S =
∮
C
u ⋅ d r
of 35 54
We can make this into two integrals:
Equation 8.3.
Equation 8.4
If in the equation (The accurate harmonic mean form):
Equation 8.5
We make the approximation
Equation 8.6
Then the Stokes form of the equation becomes
Equation 8.7
Thus we see for this approximation there are two integrals as well:
Equation 8.8
Equation 8.9
∫
1
0
∫
1
0
[
Si
B
(As − Ga) +
Ge
B
(P − Al )
]
d x d y ≈
1
Ge − Si
∫
Ge
Si
xd x
∫
1
0
∫
1
0
Si
B
(As − Ga)dydz ≈
1
3
1
(Ge − Si)
∫
Ge
Si
xd x
∫
1
0
∫
1
0
Ge
B
(P − Al )d xdz ≈
2
3
1
(Ge − Si)
∫
Ge
Si
yd y
Si
B
(As − Ga) +
Ge
B
(P − Al ) =
Ge − Si
Ge
Si
dx
x
2SiGe
Si + Ge
≈ Ge − Si
∫
1
0
∫
1
0
[
Si
B
(As − Ga) +
Ge
B
(P − Al )
]
dyd z =
∫
Ge
Si
d x
∫
1
0
∫
1
0
Si
B
(As − Ga)dydz =
1
3
∫
Ge
Si
dz
∫
1
0
∫
1
0
Ge
B
(P − Al )d ydz =
2
3
∫
Ge
Si
dz
of 36 54
For which the respective paths are
One of the double integrals on the left is evaluated in moles per grams, the other grams
per mole (0 to 1 moles per gram and 0 to 1 grams per mole).
By making the approximation
In
We have
Equation 8.10
is the differential across Si, is the differential across Ge
and is the vertical differential.
Which is Ampere’s Circuit Law
We see if written
Which is interesting because it is semiconductor elements by molar mass which are used
to make circuits.
We say (Phi) is given by
y
1
=
1
3
B
SiGa
ln(z)
y
2
=
2
3
B
Si Al
ln(z)
2SiGe
Si + Ge
≈ Ge − Si
Si(As − Ga)
B
+
Ge(P − Al )
B
=
2SiGe
Si + Ge
Si
ΔGe
ΔS
+ Ge
ΔSi
ΔS
= B
ΔSi = P − Al
ΔGe = As − Ga
ΔS = Ge − Si
∇ ×
B = μ
0
J + μ
0
ϵ
∂
E
∂t
Si
ΔGe
ΔS
= B − Ge
ΔSi
ΔS
Φ
of 37 54
and
And
=1.618
=0.618
(phi) the golden ratio conjugate. We also find
Equation 8.11
Thus since
And we have
Equation 8.12
We see and are both and c is in the Si (silicon) field wave, but for E and B fields c is the
speed of light.
a = b + c
a
b
=
b
c
Φ = a /b
ϕ = b /a
ϕ
(ϕ)ΔGe + (Φ)ΔSi = B
∇ ×
B = μ
J + μϵ
0
∂
E
∂t
Si
ΔGe
ΔS
= B − Ge
ΔSi
ΔS
ΔG e =
ΔS
Si
B −
G e
Si
ΔSi
(
∇
2
−
1
c
2
∂
2
∂t
)
E = 0
(
∇
2
−
1
c
2
∂
2
∂t
)
B = 0
c =
1
ϵ
0
μ
≈ ϕ
μ
ϵ
0
Φ
ϕ
ϵ
0
= 8.854E − 12F ⋅ m
−1
μ = 1.256E − 6H /m
G e
Si
= μ ϵ
0
of 38 54
To find the Si wave our differentials are
It is amazing how accurately we can fit these differentials with an exponential equation for the
upward increase. The equation is
This is the halfwave:
ΔS
Si
= μ
(
∇
2
−
1
ϕ
2
∂
2
∂x
)
Si = 0
(
∇
2
−
1
ϕ
2
∂
2
∂x
)
G e = 0
ΔC = N − B = 14.01 − 10.81 = 3.2
ΔSi = P − Al = 30.97 − 26.98 = 3.99
ΔGe = A s − G a = 74.92 − 69.72 = 5.2
ΔSn = Bi − In = 121.75 − 114.82 = 6.93
ΔPb = Bi − Tl = 208.98 − 204.38 = 4.6
y(x) = e
0.4x
+ 1.7
y(x) = e
2
5
x
+
17
10
of 39 54
Equation 8.3
Building Another Matrix
We pull these AI elements out of the periodic table of the elements to make an AI periodic table:
We now notice we can make a 3 by 3 matrix of it, which lends itself to to the curl of a vector field
by including biological elements carbon C (above Si):
=
=
Which results in Stokes theorem:
y(x) = e
0.4x
+ 1.7
y(x) = e
2
5
x
+
17
10
y(x) = e
B
Al
x
+
A g
Cu
B
Al
=
10.81
26.98
= 0.400667
A g
Cu
=
107.87
63.55
= 1.6974 ≈ 1.7
i
j
k
∂
∂x
∂
∂ y
∂
∂z
( − C ⋅ P)y (Si ⋅ G a)z (G e ⋅ As)y
(Ge ⋅ A s − Si ⋅ Ga)
i + (C ⋅ P)
k
[
(72.64)(74.92) − (28.09)(69.72)
]
i +
[
(12.01)(30.97)
]
k
of 40 54
Where
We then able to write this with product notation
While we have the AI BioMatrix
Which we use to formulate a similar equation. We can form another 3X3 matrix we will call the
electronics matrix:
We can remove the 5th root sign in the above equation by noticing
=(28.085)(72.64)(12.085)(107.8682)(196.9657)=
5
∫
Ge
Si
∫
Ge
Si
∇ ×
u ⋅ d
a = ex p
(
1
Ge − Si
∫
Ge
Si
ln(x)d x
)
∇ ×
u = (G e ⋅ As − Si ⋅ Ga)
i + (C ⋅ P)
k
d
a =
(
zd yd z
i + yd yd z
k
)
u = ( − C ⋅ P )y
i + (Si ⋅ G e)z
j + (G a ⋅ As)y
k
5
∫
Ge
Si
∫
Ge
Si
∇ ×
u ⋅ d
a =
n
n
∏
i=1
x
i
5
∏
i=1
x
i
= Si ⋅ Ge ⋅ Cu ⋅ A g ⋅ Au
of 41 54
Where we have substituted carbon (C=12.01) the core biological element for copper (Cu).
But since we have:
Equation 8.12
We take the ratio and have
Almost exactly 3 which is the ratio of the perimeter of regular hexagon to its diameter used to
estimate pi in ancient times by inscribing it in a circle:
Perimeter=6
Diameter=2
6/2=3
Thus we have the following equation…
Showing The Calculation using the most accurate data possible…
Ge=72.64
As=74.9216
Si=28.085
Ga=69.723
C=12.011
P=30.97376200
=
523,818, 646.5
g
5
m ol
5
∫
Ge
Si
∫
Ge
Si
( ∇ ×
u) ⋅ d
a = 170,535, 359.662(g/m ol )
5
523,818, 646.5
170,535, 359.662
= 3.0716
π = 3.141 . . .
π
∫
Ge
Si
∫
Ge
Si
( ∇ ×
u) ⋅ d
a =
5
∏
i=1
x
i
(Ge ⋅ A s − Si ⋅ Ga)
i + (C ⋅ P)
k
of 42 54
=
=154,082,837.980+16,452,521.6822=
=
(28.085)(72.64)(12.085)(107.8682)(196.9657)=
Where we have substituted carbon C=12.01 for copper Cu. We use Cu, Ag, Au because they are
the middle column of our electronics matrix, they are the finest conductors used for electrical
wire. We use C, Si, Ge because they are the middle column of our AI Biomatrix. Si and Ge are the
primary semiconductor elements used in transistor technology (Artificial Intelligence) and C is
the core element of biological life. We have
Perimeter/Diameter of regular hexagon = 3.00
[
(72.64)(74.9216) − (28.085)(69.723)
]
i +
[
(12.011)(30.97376200)
]
k
3,484 . 134569
(
g
m ol
)
2
i + 372.025855
(
g
m ol
)
2
k
∫
Ge
Si
∫
Ge
Si
(
3,484 . 134569
(
g
m ol
)
2
i + 372.025855
(
g
m ol
)
2
k
)
⋅
(
zd yd z
i + yd yd z
k
)
∫
Ge
Si
∫
Ge
Si
(
3,484 . 134569
(
g
m ol
)
2
⋅ z d zd y + 372.025855
(
g
m ol
)
2
⋅ yd z d y
)
∫
Ge
Si
3,484 . 134569
(
(72.64 − 28.085)
2
2
)
d y +
∫
Ge
Si
372.025855y ⋅ (72.64 − 28.085)d y
3458261.42924
(
g
m ol
)
4
(72.64 − 28.085) + 16575.6119695
(
g
m ol
)
3
(
(72.64 − 28.085)
2
2
)
170,535, 359.662
(
g
m ol
)
5
5
∏
i=1
x
i
= Si ⋅ Ge ⋅ C ⋅ Ag ⋅ Au
523,818, 646.5
g
5
m ol
5
523,818, 646.5
170,535, 359.662
= 3.0716
π = 3.141 . . .
of 43 54
The same value as our 3.0716 if taken at two places after the decimal.
9.0 The Basis The transformation that rotates counter-clockwise where
is given by the standard matrix
Equation 7.1
We suggest there is an aspect of Nature founded on six-fold symmetry, the example of
which we are interested in here is The Periodic Table of the Elements, because it has 18
groups which we can define by carbon, C. This because we have the following scenario:
Equations 7.2
And, we pull out the 2 and the 3 and write (Fig. 9.1)
, ,
Where , such that
which is given by
3.141 + 3.00
2
= 3.0705
T : ℝ
2
→ ℝ
2
T(
x ) = A
x
A =
(
cos(θ ) −sin(θ )
sin(θ ) cosθ
)
3 + 3 + 3 = 9
2 + 2 + 2 = 6
3 ⋅ 6 = 18
2 ⋅ 9 = 18
2 ⋅ 3 = 6
2cos
π
4
= 2
2cos
π
5
=
5 + 1
2
2cos
π
6
= 3
5 + 1
2
= Φ =
a
b
a
b
=
b
c
a = b + c
ϕ =
b
a
=
1
Φ
a
2
b
2
−
a
b
− 1 = 0
Fig. 9.1 Dividing line in
golden mean.
of 44 54
In general
Equations 7.3 ,
, ,
And these can be mapped by the matrix A onto a linear vector
space (Fig. 9.2)
=
=
f (n) = 2 cos
π
n
n = 4,5,6
π
4
= 45
∘
π /5 = 36
∘
π
6
= 30
∘
A =
(
2cos(θ ) −2sin(θ )
2sin(θ ) 2cosθ
)
A =
(
2cos(30
∘
) −2sin(30
∘
)
2sin(30
∘
) 2cos(30
∘
)
)
A
e
1
=
3 −1
1 3
(
1
0
)
= ( 3,1)
A
e
2
=
3 −1
1 3
(
0
1
)
= (−1, 3)
A =
(
2cos(36
∘
) −2sin(36
∘
)
2sin(36
∘
) 2cos(36
∘
)
)
Φ −
1
2
(5 − 5)
1
2
(5 − 5) Φ
Fig.9.2!
of 45 54
=
Our is based on the square (Fig. 9.3)
is the line . The reflection through is given by:
Equation 7.4
And our is the equilateral triangle:
A
e
1
= Φ,
1
2
(5 − 5)
A
e
2
=
1
2
(5 + 5), Φ
A =
(
2cos(45
∘
) −2sin(45
∘
)
2sin(45
∘
) 2cos(45
∘
)
)
2 − 2
2 2
A
e
1
= 2, 2
A
e
2
= − 2, 2
2cos
π
4
π
4
= 45
∘
x
2
= x
1
x
2
= x
1
A =
(
0 1
1 0
)
2cos
π
6
Fig. 9.3
of 46 54
To transform the square into the equilateral triangle we expand the square of base
with the matrix
Equation 7.5
And we see becomes and we have added half the square to itself. (Fig. 9.4)
Or, better we can use the contraction
We draw in the diagonal of the the half-square and reassemble the two half-triangles
into an equilateral triangle (Fig. 9.8). To get we take the half square and draw in the
circle of radius 1/2. (Fig. 9.5) We have
e
1
A =
(
3/2 0
0 1
)(
1
0
)
=
3
2
,1
e
1
3/2
A =
(
1/2 0
0 1
)(
1
0
)
=
1
2
Φ
1
2
2
+ 1
2
=
1
4
+
4
4
=
5
2
5
2
+
1
2
=
5 + 1
2
= Φ
Fig. 9.4
of 47 54
Thus we see the periodic table is 18 groups (Fig. 9.6). Carbon is in group 14. We have
18-14=4 valence electrons. Hydrogen is neither a metal or a non-metal but ionizes like a
metal by losing one electron becoming and carbon being means it needs 4
positive ions to be neutral meaning it combines with 4 hydrogens to each C, or with two
hydrogens to a C and a C in long chains (hydrocarbons) which form the Skeltons of
organic compounds in life chemistry (Fig. 9.7) .
H
+
C
4−
Fig. 9.5
Fig. 9.6
Fig. 9.7
Fig. 9.8
of 48 54
That is:
is phenomenal because It allows multiplication between degrees and seconds to output
our fundamental ratios ( ). We see in the following wave:
, ,
Where t=1 second is carbon yielding:
And, t=6 seconds is hydrogen yielding:
10.0 Solar System
In so far as
relates carbon=1second to the Earth-Moon-Sun orbital mechanics and to the radius of a
proton through six-fold symmetry:
can we bring in our relationships
,
,
1
6protons
⋅
1
α
2
m
p
h4πr
2
p
Gc
= 1.00second
2, 3, . . .
A = A
0
cos(θt)
A
0
= 1
θ = 30
∘
,60
∘
,45
∘
A(60
∘
) = cos(60
∘
⋅ 1s) = 0.5
A(30
∘
) = cos(30
∘
⋅ 1s) = 3 /2
A(45
∘
) = cos(45
∘
⋅ 1s) = 2 /2
A(60
∘
) = cos(60
∘
⋅ 6s) = 1
A(30
∘
) = cos(30
∘
⋅ 6s) = − 1
A(45
∘
) = cos(45
∘
⋅ 6s) = 0
1
α
2
m
p
h4πr
2
p
Gc
≈ 6 ⋅
KEof Moon
KEof Ear th
⋅ Ear th Day ≈ (6)1secon d
1
α
2
m
p
h4πr
2
p
Gc
= 6
f (n) = 2cos
π
n
n = 2,3,4,5.6.,...
f (2) = 2
f (6) = 3
of 49 54
to expand the Earth-Moon-Sun orbital mechanics to the size of the Solar System? We
consider:
Equations 10.1
We move down from carbon in the periodic table to silicon (Si) and down from there to
germanium (Ge). Their densities are Si=2.33 g/cm3 and Ge=5.323 g/cm3. We have
0.21Si+0.78Ge=4.64124 g/cm3. Consider this the starting point for the density of a thin
disc decreasing linearly from the Sun to Pluto (49.5 AU=7.4E14cm).
Equation 10.2
=
The sum of the masses of the planets is 2.668E30 grams. The accuracy is:
We have the radius of the solar system is given by:
Equation 10.3
Interestingly, the relative abundances of Nitrogen and oxygen are 0.78 and 0.21 in the
Earth air. In fact, the molar mass of air as a mixture is:
Interestingly as well, by molar mass
6
∫
3
2
1
t
2
dt = 6
(
1
2
−
1
3
)
= 0.78
∫
2
3
cos
−1
(x /2)d x =
1
6
(
3π − 6
)
=
π − 4
2 2 = 0.21
= 0.21
M =
∫
2π
0
∫
R
0
ρ
0
(
1 −
r
R
)
rdrdθ =
πρ
0
R
2
3
π(4.64124)(7.4E14)
2
3
= 2.661E 30g
2.66
2.668
100 = 99.736
R
s
=
3M
p
π(0.78Ge + 0.21Si )
air = 0.78N
2
+ 0.21O
2
= 29.0g/mol
of 50 54
Carbon equals one second is the radius of a proton:
Equation 10.4
The experimental radius of a proton is:
In our integral:
where did come from? We had Equations 3
,
We write and get and
11.0 Natural Abundances
If we can explain the Earth-Moon-Sun orbital mechanics, the radius of the solar system,
and the radius of the proton, can we predict the relative abundances of the primordial
elements created at the beginning of the Universe - hydrogen and helium - from which
all of the elements were made by stars in nucleosynthesis?
We consider a gaussian wave packet at t=0:
air
H
2
O
≈ Φ
r
p
=
18
3
α
2
m
p
Gc
4πh
= 8.288587 × 10
−16
m = 0.829f m
r
p
= 0.833
±
0.014 f m
∫
2
3
cos
−1
(x /2)d x =
1
6
(
3π − 6
)
= 0.21
cos
−1
(x /2)
f (n) = 2cos
π
n
n = 4,5,6
x = 2cosθ
θ = cos
−1
(x /2)
cos
−1
( 2/2) = 45
∘
=
π
4
cos
−1
( 3/2) = 30
∘
=
π
6
ψ (x,0) = Ae
−
x
2
2d
2
of 51 54
We say that d which in quantum mechanics would be the delocalization length when
squared is
!
We write the wave packet as a Fourier transform:!
!
We use the identity that gives the integral of a quadratic:!
!
!
Calculate the Gaussian integral of dp:!
and. !
Equation 11.1!
!
!
(
Si − C
C
)
2
=
16
9
ψ (x,0) = Ae
−
x
2
2d
2
=
∫
dp
2π ℏ
ϕ
p
e
i
h
px
∫
∞
−∞
e
−α
2
x+βx
d x =
π
α
e
β
2
4α
ψ (x, t) =
∫
dp ⋅ e
p
2
d
2
2ℏ
2
⋅ e
−
i
h
( px−
p
2
2m
t)
α =
d
2
2ℏ
2
+
it
2mℏ
β =
i x
ℏ
ψ
2
= exp
[
−
x
2
d
2
⋅
1
1 + t
2
/τ
2
]
τ =
m d
2
ℏ
ψ
2
= exp −
C
2
x
2
(Si − C )
2
⋅
1
1 +
[
ℏC
2
m(S i − C)
2
]
2
t
2
ψ
2
= exp −
9
16
x
2
⋅
1
1 +
ℏ
2
81
m
2
256
t
2
of 52 54
For Hydrogen (t=6 seconds):
Equation 11.2
=74%
For Helium (t=3 seconds):
Equation 11.3
=26%
This is in close agreement with what we observe. The Universe is 74% hydrogen and
24% helium, The remaining 2% is comprised by the rest of the elements.
ψ
2
= (1)exp −
9
16
(1proton)
2
⋅
1
1 +
(0.075)81
(1)256
(6seconds)
2
ψ
2
= (2)exp −
9
16
(2proton)
2
⋅
1
1 +
(0.075)81
(4)256
(3seconds)
2
of 53 54
Appendix 1
To find the translational kinetic energy of the moon:
Distance from earth: 3.85E8m
Orbital period:
T=27.32 days=2.36E6seconds
v=1.025E3m/s~1000m/s
Mass: 7.34767E22kg
Use
E=3.67E28 Joules
To find the translational kinetic energy of the earth:
Distance from Sun: 1AU=1.496E11m
Orbital period: 1 year=3.1558E7 seconds
v=2.9785E4m/s
E=2.649E33 Joules
Earth day=(24)(60)(60)=86,400 seconds.
2π (3.85E8m) = 2.419E 9m
E =
1
2
mv
2
2π (1.496E11m) = 9.399E11m
of 54 54
The Author
